The Encyclopedia of Triangle Centers lists X(175) as the isoperimetric point and that its trilinear coordinates are $-1 + \sec \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.
I have, however not been able to find a proof of this. Can someone provide the proof?
We just need to read a few lines of the Wikipedia page about the isoperimetric point to be convinced of the following fact: given the circles $\Gamma_A,\Gamma_B,\Gamma_C$ centered at $A,B,C$ and going through the projections of the incenter on the triangle sides, there are two circles simultaneously tangent to $\Gamma_A,\Gamma_B,\Gamma_C$. If $\Gamma_A,\Gamma_B,\Gamma_C$ lie inside the larger one (outer Soddy circle), the centre of the outer Soddy circle is a isoperimetric point. It is not difficult to show that the isoperimetric point, if exists, is unique, then we just need a bit of trigonometry$^{(*)}$ and/or circle inversions to locate the centre of the outer Soddy circle, i.e. to find its trilinear coordinates.
Guide to $(*)$: perform a circular inversion with respect to a circle centered at the tangency point $P$ of $\Gamma_C$ and $\Gamma_B$ having radius $R^2=(s-c)(s-c)$. That brings $\Gamma_B$ and $\Gamma_C$ into two parallel lines and $\Gamma_A$ into a circle tangent to both lines. The involved lengths in the new configuration just depend on $a$ and $r\cot\frac{B}{2}\cot\frac{C}{2}$. Translate the image of $\Gamma_A$ to find the inverse of the outer Soddy circle and consider a diameter of that circle that goes through $P$. Take the inverse of the endpoints of such a diameter to get a diameter of the outer Soddy circle, then compute the distance of the midpoint of such a diameter from the $BC$-side.
